Electric
Electric
Fields
Fields
andand
Dipole
Gauss’s Law
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
1
Announcements
!  Homework set 1 is due 1/20
!  Lecture notes: linked from lon-capa, or directly at
http://www.pa.msu.edu/~schwier/courses/2014SpringPhy184/
!  Section 1 lecture notes:
http://www.pa.msu.edu/~nagy_t/phy184/lecturenotes.html
!  Strosacker learning center schedule – BPS 1248
•
•
•
•
Mo: 10am – noon, 1pm – 9pm
Tue: noon – 6pm
We: noon – 2pm
Th: 10am – 1pm, 2pm – 9pm
!  Clicker:
•  If you didn’t enter it on lon-capa, then email me your clicker
number
•  Clickers only work for the section in which you are registered
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 21
2
The Electric Field
!  The electric field is defined at any point in space as the net
electric force on a charge, divided by that charge
 

F (r )
E (r ) =
q
!  Electric field lines
start at positive
charges and end at
negative charges
!  Electric field line density:
+
Test charge q
•  High density of field lines – large field
•  Low density of field lines – small field
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
4
The Electric Field and force
!  The electric force on a charge is parallel or antiparallel to the
electric field at that point
E
F
positive charge
E
F
negative charge
!  The electric force is
 

F ( r ) = qE ( r )
January 14, 2014
F= qE
Physics for Scientists & Engineers 2, Chapter 22
5
Superposition of electric fields
!  The total electric field at any point is the vector sum of all
individual electric fields


Enet = ∑ Ei
i
!  Note that this means adding the x-, y-, and z- components
separately
Enet ,X = ∑ Ei ,X
i
Enet ,Y = ∑ Ei ,Y
i
Enet ,Z = ∑ Ei ,Z
i
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 21
6
Electric Field from 4 Point Charges
PROBLEM
!  We have four charges
•
•
•
•
q1 = 10.0 nC
q2 = -20.0 nC
q3 = 20.0 nC
q4 = -10.0 nC
!  These charges form a square of
edge length 5.00 cm
!  What electric field do the particles
produce at the square center?
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
8
Problem solving strategy
!  Use this approach to solve problems, in particular if at first you have no clue.
Step 1
Think
Step 2
Sketch
Recognize the problem
What’s going on?
Think
Describe the problem
in terms of the field
What does this have to do with…?
Sketch
Step 3
Research
Plan a solution
How do I get out of this?
Research
Step 4
Execute
Step 5
Double-check
1/14/14
Execute the plan
Let’s get an answer!
Simplify, Calculate, Round
Evaluate the solution
Can this be true?
!  Draw a picture
!  Phrase the question in your own
words
!  Relate the question to something
you just learned
!  Identify physics quantities, forces,
fields, potentials,…
!  Find a physics principle
(symmetry, conservation, …)
!  Write down the equations
!  Solve equations, starting with
intermediate steps
!  Check units, order-of-magnitude,
insert into original question, …
Double-check
Physics for Scientists & Engineers 2
9
Electric Field from 4 Point Charges
PROBLEM
!  We have four charges
•
•
•
•
q1 = 10.0 nC
q2 = -20.0 nC
q3 = 20.0 nC
q4 = -10.0 nC
!  These charges form a square of
edge length 5.00 cm
!  What electric field do the particles
produce at the square center?
SOLUTION
THINK
!  Each of the four charges produces a field at the center
!  We can use the principle of superposition
•  The electric field at the center of the square is the vector sum of the
electric field from each charge
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
10
Electric Field from 4 Point Charges
SKETCH
!  We define an x-y coordinate system
•  We place q2 and q3 on the x axis
•  We place q1 and q4 on the y axis
•  The center of the square is at
x=y=0
•  The sides of the square are
a = 5.00 cm
•  The distance from each charge to
the center is r
RESEARCH
!  The electric field at the center of the square is given by the
principle of superposition

   
Ecenter = E1 + E2 + E3 + E4
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
11
Electric Field from 4 Point Charges
!  We can write the field in terms of the x- and y-components
Ecenter,x = E1,x + E2,x + E3,x + E4,x
Ecenter,y = E1,y + E2,y + E3,y + E4,y
!  All four charges are a distance r from the center
2a a
r=
=
2
2
!  The electric field component from each charge at the center
of the square is given by
qi
r2
SIMPLIFY
Ei = k
!  Let’s start with the x-components
Ecenter,x = E1,x + E2,x + E3,x + E4,x
January 14, 2014
q2
q3 k
= k 2 − k 2 = 2 ( q2 − q3 )
r
r
r
Physics for Scientists & Engineers 2, Chapter 22
12
Electric Field from 4 Point Charges
!  Now the y-components
Ecenter,y = E1,y + E2,y + E3,y + E4,y = k
q1
q4 k
−
k
= 2 ( q1 − q4 )
2
2
r
r
r
!  The magnitude of the electric field at the center is
2
2
Ecenter = Ecenter
,x + Ecenter ,y
Ecenter
k
= 2
r
(q2 − q3 ) + (q1 − q4 )
2
2
2k
= 2
a
(q2 − q3 ) + (q1 − q4 )
2
2
⎛ 2 a2 ⎞
⎜⎝ r = 2 ⎟⎠
!  The angle is given by
⎛ k
⎞
q
−
q
(
)
1
4
2
⎛ Ecenter,y ⎞
⎟
−1
−1 ⎜ r
−1 ⎛ q1 − q4 ⎞
θ = tan ⎜
= tan ⎜
= tan ⎜
⎟
⎟⎠
⎟
k
q
−
q
⎝
⎝ Ecenter,x ⎠
2
3
⎜⎝ 2 ( q2 − q3 ) ⎟⎠
r
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
13
Electric Field from 4 Point Charges
CALCULATE
!  Putting in our numerical values we get
2k
2
2
q
−
q
+
q
−
q
(
( 1 4)
2
3)
2
a
(q2 − q3 ) = ( −20.0 nC ) − ( 20.0 nC ) = −40.0 nC = −40.0 ⋅10−9 C
Ecenter =
(q1 − q4 ) = (10.0 nC ) − ( −10.0 nC ) = 20.0 nC = 20.0 ⋅10−9 C
Ecenter =
2
⎛
9 Nm ⎞
2 ⎜ 8.99 ×10
C 2 ⎟⎠
⎝
(0.0500 m )
2
( −40.0 ⋅10
−9
C ) + ( 20.0 ⋅10 C )
2
−9
2
Ecenter = 321636.0179 N/C
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
14
Electric Field from 4 Point Charges
!  For the angle we get
⎛ q1 − q4 ⎞
θ = tan ⎜
⎝ q − q ⎟⎠
−1
2
3
⎛ 10.0 − (−10.0) ⎞
−1
θ = tan ⎜
=
tan
−0.500 )
(
⎟
⎝ −20.0 − 20.0 ⎠
−1
θ = −26.56505118°
ROUND
!  We round our results to three significant figures
Ecenter = 3.22 ⋅10 5 N/C
θ = −26.6°
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
15
Electric Field from 4 Point Charges
DOUBLE-CHECK
!  What about the angle?

E3

E4
January 14, 2014
x

E2

E1
θ = −26.6°
Physics for Scientists & Engineers 2, Chapter 22

Ecenter
16
Electric Field from an Electric Dipole
!  A system of two oppositely charged point particles is called
an electric dipole
!  The vector sum of the electric field from the two charges
gives the electric field of the dipole
!  We have shown the electric field lines from a dipole
!  We will derive a general
expression good
anywhere along the
dashed line and then get
an expression for the
electric field a long
distance away from the
dipole
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
18
Electric Field from an Electric Dipole
!  Start with two charges on the x-axis a distance d apart
•  Put -q at x = -d/2
•  Put +q at x = +d/2
!  Calculate the electric field at a point P a distance x from the
origin
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
19
Electric Field from an Electric Dipole
!  The electric field at any point x is the sum of
the electric fields from +q and –q
1 q
1 −q
E = E+ + E− =
+
2
4πε 0 r+ 4πε 0 r−2
!  Replacing r+ and r- we get the electric field
everywhere on the x-axis (except for x = ±d/2)
⎤
q ⎡
1
1
E=
⎢
2 −
2 ⎥
1
1
4πε 0 ⎢⎣ ( x − 2 d ) ( x + 2 d ) ⎥⎦
!  Interesting limit far away along the positive xaxis (x >> d)
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
20
Electric Field from an Electric Dipole
!  We can rewrite our result as
−2
−2
⎡
q
d ⎞
d ⎞ ⎤
⎛
⎛
E=
1 − ⎟ − ⎜1 + ⎟ ⎥
2 ⎢⎜
⎝
⎝ 2x ⎠ ⎦
4πε 0 x ⎣
2x ⎠
!  We can use a binomial expansion or a Taylor expansion
d
n
α=
, (1 ± α ) ≈ 1 ± nα + ...
2x
d
−2
(1 − α ) = 1 − (−2)α ... = 1 + 2 + ...
2x
d
−2
(1 + α ) = 1 + (−2)α ... = 1 − 2 + ...
2x
f (z) = (1 ± z)n
∞
f (z) = ∑
n=0
f (n) (z = 0)
(z − 0)n
n!
For small z=d/2x, stop after n=1
!  So we can write
q
E≈
4πε 0 x 2
January 14, 2014
q
⎡⎛ d ⎞ ⎛ d ⎞ ⎤
⎢⎜⎝ 1 + x ⎟⎠ − ⎜⎝ 1 − x ⎟⎠ ⎥ = 4πε x 2
⎣
⎦
0
qd
⎡ 2d ⎤
⎢⎣ x ⎥⎦ = 2πε x 3
0
Physics for Scientists & Engineers 2, Chapter 22
21
Definition of Electric Dipole Moment
!  We define the electric dipole moment as a vector that
points from the negative charge to the positive charge


p = qd
•  p is the magnitude of the dipole moment
•  q is the magnitude of one of the opposite charges
•  d is the distance between the charges
!  Using this definition we can write the electric field far away
from an electric dipole as
p
E=
2πε 0 x 3
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
22
Electric Dipole Moment of Water
!  Chemistry reminder
•  the H2O molecule
!  The distribution of electric charge in
an H2O molecule is non-uniform
!  The more electronegative oxygen
atom attracts electrons from the
hydrogen atoms
!  Thus, the oxygen atom acquires a partial negative charge
and the hydrogen atoms acquire a partial positive charge
!  The water molecule is “polarized”
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
23
Electric Dipole Moment of Water
PROBLEM
!  Suppose we approximate the water molecules as two
positive charges located at the center of the hydrogen atoms
and two negative charges located at the center of the oxygen
atom
!  What is the electric dipole moment of a water molecule?
SOLUTION
!  The center of charge of the two positive
charges is located exactly halfway between the
centers of the hydrogen atoms
!  The distance between the positive and negative
charge centers is
⎛θ⎞
d = Δr cos ⎜ ⎟ = (10−10 m ) cos(52.5°) = 0.6 ⋅10−10 m
⎝ 2⎠
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
24
Electric Dipole Moment of Water
!  This distance times the transferred charge, q = 2e,
is the magnitude of the dipole moment of water
p = 2ed = 2 (1.6⋅10−19 C )( 0.6⋅10−10 m ) = 2⋅10−29 C m
!  This oversimplified calculation comes fairly close
to the measured value of the electric dipole
moment of water of 6.2·10-30 C m
!  The fact that the real dipole moment of water is less than
our calculated result indicates that the two electrons of the
hydrogen atoms are not pulled all the way to the oxygen, but
only one-third of the way
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 22
25
Demo: electric force on dipole
!  Demo: electric force on conducting end of stick balanced at
mid-point
!  Demo: electric force on water
•  Force due to dipole moment of water
!  Demo: electric force on wooden end of stick balanced at
mid-point
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 21
26
Explanation of the Demo
!  The presence of the charged stick polarizes the atoms even
on the wooden side of the stick
January 14, 2014
Physics for Scientists & Engineers 2, Chapter 21
27